Suppose that I have n independent Bernoulli random variables

Suppose that I have n independent Bernoulli random variables, X1, X2, ..., Xn. Recall that a Bernoulli distribution random variable takes the value 0 with probability 1-p and the value 1 with probability p. We showed in class thatthemeanandvarianceofanyoneoftheX’s are = p and ^2=p(1p). Thatis E(Xi) = p,i=1, ..., n and V ar(X i) = p(1 p), i = 1, ..., n . Note that your answers below will depend on “n” and “p”, and should agree with the mean and variance of a Binomial (n,p) distribution, since the number of successes can be written as the sum of n Bernoulli random variables.

Calculate a formula for the expected value of X1 + X2 + ... + Xn .

Calculate a formula for the variance of X1 + X2 + ... + Xn .

Solution

X1,X2 , X3 ...Xn are Bernoulli random variables and we know that

E(X1) = 0*(1-p) + 1*p = p

Var(X1) = 0^2* (1-p) + 1^2 * p -(p^2) = p-p^2 = p(1-p)

E(X1+X2+X3+...+Xn) = E(X1)+E(X2)+E(X3)+....+E(Xn) = p+p+p+...+p (n times) = np

Var(X1+X2+X3+...+Xn) = Var(X1)+Var(X2)+....+Var(Xn)+ 2cov(X1,X2)+2cov(X1,X3)+....+2cvv(Xn-1,Xn)

X1,X2,....Xn are independent Bernoulli random variables, so Cov(Xi,Xj) = 0

Therefore

Var(X1+X2+X3+...+Xn) = Var(X1)+Var(X2)+....+Var(Xn)= p(1-p)+p(1-p)+....+p(1-p) ( n times)

= np(1-p) = npq ( where q =1-p)

Hope this helps you